Question: Aditya's dog routinely eats Aditya's leftovers, which vary seasonally. As a result, his weight fluctuates throughout the year. The dog's weight $W(t)$ (in $\text{kg}$ ) as a function of time $t$ (in days) over the course of a year can be modeled by a sinusoidal expression of the form $a\cdot\cos(b\cdot t)+d$. At $t=0$, the start of the year, he is at his maximum weight of $9.1\text{ kg}$. One-quarter of the year later, when $t=91.25$, he is at his average weight of $8.2\text{ kg}$. Find $W(t)$. $\textit{t}$ should be in radians. $W(t) = $
Answer: The strategy First, we should convert the given information about the real-world context into mathematical terms of the sinusoidal function and its graph. Then, we should use the given information to find the amplitude, midline, and period of the function's graph. Finally, we should find $a$, $b$, and $d$ in the expression $a\cos(b\cdot t)+d$ by considering the features we found. Converting the given information into mathematical terms At $t=0$, Aditya's dog weighs $9.1\text{ kg}$. This means the graph of the function passes through $(0,9.1)$. We are given that this is his maximum weight, which corresponds to a maximum of the graph. $91.25$ days later (which means $t=91.25$ ) he weighs $8.2\text{ kg}$. This corresponds to the point $(91.25,8.2)$. We are given that this is his average weight, which corresponds to the midline of the graph. In conclusion, the graph has a maximum point at $(0,9.1)$ and then intersects its midline at $(91.25,8.2)$. Determining the amplitude, midline, and period The midline intersection is at $y={8.2}$, so this is the midline. The maximum point is $0.9$ units above the midline, so the amplitude is ${0.9}$. The maximum point is $91.25$ units to the left of the midline intersection, so the period is $4\cdot 91.25={365}$. [Why did we multiply by 4?] Determining the parameters in $a\cos(b\cdot t)+d$ Since the maximum at $t=0$ is followed by a midline intersection, we know that $a>0$. [How do we know that?] The amplitude is ${0.9}$, so $|a|={0.9}$. Since $a>0$, we can conclude that $a=0.9$. The midline is $y={8.2}$, so $d=8.2$. The period is ${365}$, so $b=\dfrac{2\pi}{{365}}$. The answer $W(t)=0.9\cos\left(\dfrac{2\pi}{365}t\right)+8.2$